Class 10 Science · Chapter 11

Electricity

218 Class 10 Science questions on Electricity from the NCERT textbook, NCERT Exemplar and CBSE board papers 2023–2026, each with its answer and a step-by-step solution.

CBSE Class 10 Science previous year papers, with answers →

Questions · page 2 of 11

1. Q21 NCERT · Exercises
   Explain the following.

   1. (a)Why is ​the tungsten used almost exclusively for filament of ​electric lamps?
   2. (b)​Why are the ​conductors of ​electric ​heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
   3. (c)Why is the series ​arrangement not used for domestic circuits?
   4. (d)How ​does the resistance of a wire vary with its area of cross-section?
   5. (e)Why are copper and aluminium wires usually employed for electricity transmission?
   Show answer

   Tungsten has a high melting point; alloys have high resistivity; series wiring shares voltage; $R\propto 1/A$; copper and aluminium lose little heat.

   (a) Tungsten has a very high melting point (about $3380^{\circ }\text{C}$), well above the roughly $2500^{\circ }\text{C}$ at which a filament glows, together with a high resistivity; it retains the heat produced and does not oxidise or vaporise readily, so it can be kept white-hot without melting away. (b) An alloy such as nichrome has a much higher resistivity than the pure metals in it, so it develops far more heat for the same current ($H=I^{2}Rt$), and it does not oxidise or burn readily at high temperature, so the element lasts. (c) In series the same current would pass through every appliance, the supply voltage would be divided so none would get the full $220\text{V}$, the total resistance would be large so the current would be small, and one appliance failing or being switched off would break the circuit and stop all the others; appliances are therefore connected in parallel. (d) Inversely — $R=\rho l/A$, so for a given material and length the resistance is inversely proportional to the area of cross-section: doubling the area halves the resistance. (e) Copper and aluminium have very low resistivity, so the heat lost in the line, $I^{2}Rt$, is small; they are also ductile, strong enough, and far cheaper and more plentiful than silver, with aluminium's lightness suiting long overhead lines.

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2. Q22 NCERT · Intext Questions (Page 174)
   Name ​a device ​that ​helps to ​maintain ​a potential difference ​across ​a conductor.
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   An electric cell, or a battery of cells, maintains the potential difference.

   An electric cell — or a battery, which is a combination of two or more cells — maintains the potential difference across a conductor. The chemical action inside the cell generates a potential difference across its terminals and keeps it up while current is drawn, spending the chemical energy stored in the cell. Any other source of electrical energy, such as a d.c. power supply or a generator, does the same job.

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3. Q23 NCERT · Intext Questions (Page 174)
   What is ​meant by saying ​that ​the potential ​difference ​between two ​points is 1 ​V?
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   It means 1 joule of work is done in moving 1 coulomb of charge between the two points.

   It means that $1$ joule of work is done in moving a charge of $1$ coulomb from one of the two points to the other. In symbols, $V=W/Q$, so $1\text{V}=1\text{J}/1\text{C}=1{\text{J C}}^{-1}$: every coulomb that passes between those two points gains or loses one joule of energy.

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4. Q24 NCERT · Intext Questions (Page 174)
   How much energy is ​given to ​each ​coulomb ​of charge passing ​through a 6 ​V ​battery?
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   $W=VQ=6$ J: each coulomb passing through the 6 V battery is given 6 joule of energy.

   $W=VQ=6\text{V}\times 1\text{C}=6$ J. Each coulomb of charge passing through the $6$ V battery is given $6$ joule of energy — that is exactly what the rating of $6$ V means, since $1\text{V}=1{\text{J C}}^{-1}$.

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5. Q25 NCERT · Intext Questions (Page 181)
   On ​what ​factors ​does ​the resistance of a conductor ​depend?
   Show answer

   It depends on length, area of cross-section, nature of the material (resistivity) and temperature: $R=\rho \frac{l}{A}$.

   The resistance of a conductor depends on: (i) its length $l$ — resistance is directly proportional to length, $R\propto l$; (ii) its area of cross-section $A$ — resistance is inversely proportional to it, $R\propto 1/A$, so a thicker conductor has less resistance; (iii) the nature of its material, entering through the resistivity $\rho$; and (iv) its temperature — both $R$ and $\rho$ change when the conductor is heated, which is why Ohm's law is stated for a fixed temperature. The first three are collected in $R=\rho l/A$.

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6. Q26 NCERT · Intext Questions (Page 181)
   Will current flow more easily through ​a ​thick wire ​or a thin wire of the same material, when ​connected to ​the same source? Why?
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   The thick wire, because its larger cross-section gives lower resistance, so more current flows at the same potential difference.

   Current flows more easily through the thick wire. Both wires are of the same material and the same length, so $\rho$ and $l$ are fixed and $R=\rho l/A$ leaves the resistance depending only on the area of cross-section, inversely: $R\propto 1/A$. The thick wire has the larger area, so it has the smaller resistance. The same source puts the same potential difference $V$ across each, so by Ohm's law $I=V/R$ the wire with the smaller resistance carries the larger current. Hence the thick wire draws more current — it offers an easier path.

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7. Q27 NCERT · Intext Questions (Page 181)
   Let ​the ​resistance of ​an electrical component ​remains constant while the potential difference across the two ​ends of the ​component decreases to half of its former value. What change ​will occur in the current through it?
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   The current is also halved, since $I=V/R$ and $R$ is constant.

   The current is also halved. With the resistance constant, Ohm's law $I=V/R$ makes the current directly proportional to the potential difference, so replacing $V$ by $V/2$ gives $I^{\prime }=(V/2)/R=I/2$: the new current is half the former current.

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8. Q28 NCERT · Intext Questions (Page 181)
   Why ​are ​coils of electric toasters ​and electric irons ​made ​of an alloy rather than a ​pure metal?
   Show answer

   An alloy has much higher resistivity, so it heats more for the same current, and it does not oxidise readily when red hot.

   Because an alloy suits a heating element in two ways that a pure metal does not. First, the resistivity of an alloy is generally much higher than that of its constituent metals — nichrome is about $100\times 10^{-6}\Omega \text{m}$ against about $10\times 10^{-8}\Omega \text{m}$ for iron or nickel — so a coil of ordinary length and thickness has a large resistance and, by $H=I^{2}Rt$, gives out a great deal of heat for the current drawn. Second, alloys do not oxidise (burn) readily at high temperatures, so the coil can be kept red hot in air for hours without corroding away. A pure-metal coil of the same size would hardly heat up and would soon oxidise and break.

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9. Q29 NCERT · Intext Questions (Page 181)
   Use the ​data in Table ​11.2 to answer the following –

   1. (a)​Which among ​iron ​and mercury is a better conductor?
   2. (b)Which ​material is the best ​conductor?
   Show answer

   (a) Iron, as its resistivity is lower than mercury's; (b) silver, which has the lowest resistivity in the table.

   (a) Iron is the better conductor. Table 11.2 gives ${\rho }_{\text{Fe}}=10.0\times 10^{-8}\Omega \text{m}$ against ${\rho }_{\text{Hg}}=94.0\times 10^{-8}\Omega \text{m}$; since $R=\rho l/A$, the smaller resistivity means the smaller resistance for identical wires, and iron's is smaller by a factor of about $9.4$. (b) Silver is the best conductor, with the lowest resistivity in the table, ${\rho }_{\text{Ag}}=1.60\times 10^{-8}\Omega \text{m}$ — just below copper's $1.62\times 10^{-8}\Omega \text{m}$.

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10. Q30 NCERT · Intext Questions (Page 185)
    Draw a schematic diagram of ​a circuit consisting of a ​battery of three cells of 2 V each, ​a ​5 Ω ​resistor, an 8 Ω resistor, and a 12 Ω ​resistor, and ​a ​plug key, all connected in series.
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    All parts form one series loop, so the same current flows through the $5\Omega$, $8\Omega$ and $12\Omega$ resistors and the key.

    The circuit is a single closed loop, drawn in the figure. Going round it: the battery of three $2$ V cells, drawn as three long-plate-and-short-plate pairs all facing the same way and giving $2+2+2=6$ V; then the $5\Omega$ resistor, the $8\Omega$ resistor and the $12\Omega$ resistor, each shown as a zig-zag and labelled with its value; then the plug key, drawn closed as a small circle with a dot; and plain connecting wires joining them end to end with no branch anywhere. Since everything is in series there is only one path, so the same current flows through all three resistors and the key, leaving the positive terminal of the battery and returning to its negative terminal.

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11. Q31 NCERT · Intext Questions (Page 185)
    Redraw the circuit of ​Question 1, putting in ​an ammeter to ​measure the ​current ​through the resistors and a voltmeter to measure the potential difference ​across the 12 Ω ​resistor. What would ​be the readings in the ammeter and the voltmeter?
    Show answer

    The ammeter reads 0.24 A and the voltmeter across the $12\Omega$ resistor reads 2.88 V.

    In the redrawn circuit the ammeter is cut into the loop itself, in series with the resistors so that the whole current passes through it, and the voltmeter is joined in parallel across the two ends of the $12\Omega$ resistor; each meter has its $+$ terminal on the side the current enters. The readings are: total resistance $R_s=5+8+12=25\Omega$; battery $V=3\times 2=6$ V; ammeter reading $I=V/R_s=6/25=0.24$ A; voltmeter reading $V_{12}=I{R}_3=0.24\times 12=2.88$ V. So the ammeter reads $0.24$ A and the voltmeter reads $2.88$ V.

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12. Q32 NCERT · Intext Questions (Page 188)
    Judge the equivalent resistance when ​the following are connected ​in ​parallel ​–

    1. (a)$1\Omega$ and $10^6\Omega$, ​
    2. (b)$1\Omega$ and $10^3\Omega$, and $10^6\Omega$.
    Show answer

    The equivalent resistance is about $1\Omega$ in (a) and $0.999\Omega$ in (b), just below the smallest resistance of each set.

    (a) $\frac{1}{R_p}=\frac{1}{1}+\frac{1}{10^6}=1.000001{\Omega }^{-1}$, so $R_p=0.999999\Omega \approx 1\Omega$; (b) $\frac{1}{R_p}=\frac{1}{1}+\frac{1}{10^3}+\frac{1}{10^6}=1.001001{\Omega }^{-1}$, so $R_p\approx 0.999\Omega$. In both cases the equivalent resistance is a little smaller than $1\Omega$, the smallest resistance in the combination: a very large resistance placed in parallel carries almost no current and so hardly changes the result.

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13. Q33 NCERT · Intext Questions (Page 188)
    An electric lamp ​of $100\Omega$, ​a toaster ​of resistance $50\Omega$, and a water filter of resistance $500\Omega$ ​are connected in parallel to a $220$ V source. ​What is the resistance of an electric iron connected to the same source that takes as ​much current ​as all three ​appliances, and what is the current through it?
    Show answer

    $R=31.25\Omega$ and $I=7.04\text{A}$: the iron needs the equivalent resistance of the three appliances in parallel.

    The iron must have the equivalent resistance of the three appliances in parallel: $\frac{1}{R}=\frac{1}{100}+\frac{1}{50}+\frac{1}{500}=\frac{16}{500}$, so $R=31.25\Omega$. The current through it is $I=\frac{V}{R}=\frac{220}{31.25}=7.04$ A.

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14. Q34 NCERT · Intext Questions (Page 188)
    What are ​the advantages ​of ​connecting ​electrical devices in parallel with the battery instead of connecting ​them in series?
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    In parallel each device gets the full voltage, draws its own current, keeps working if another fails, and the total resistance falls.

    Connecting devices in parallel with the battery has four advantages. (i) Each device gets the full battery voltage, because every branch is joined straight across the terminals; in series the voltage is divided among the devices, so none of them works at its rated voltage. (ii) Each device draws its own current $I=V/R$, decided by its own resistance, so appliances needing very different currents — a bulb and a heater, for instance — can run on the same supply; a series circuit forces one common current through all of them. (iii) If one device fails, only its own branch is broken and the rest keep working, whereas a break anywhere in a series circuit stops every device. Each branch can also have its own switch, so devices can be operated independently. (iv) The total resistance of a parallel combination is less than the smallest branch resistance, so the circuit can draw the larger total current that several appliances together need.

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15. Q35 NCERT · Intext Questions (Page 188)
    How ​can three resistors of resistances $2\Omega$, $3\Omega$, ​and $6\Omega$ be connected to give a total ​resistance ​of ​

    1. (a)$4\Omega$, ​
    2. (b)$1\Omega$?
    Show answer

    (a) $3\Omega$ and $6\Omega$ in parallel, in series with $2\Omega$, give $4\Omega$; (b) all three in parallel give $1\Omega$.

    (a) Connect the $3\Omega$ and $6\Omega$ resistors in parallel and put this combination in series with the $2\Omega$ resistor. The parallel pair gives $\frac{1}{R^{\prime }}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$, so $R^{\prime }=2\Omega$, and the total is $R=2+2=4\Omega$. (b) Connect all three resistors in parallel: $\frac{1}{R}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$, so $R=1\Omega$.

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16. Q36 NCERT · Intext Questions (Page 188)
    What is

    1. (a)​the highest, ​
    2. (b)the lowest ​total ​resistance ​that can be ​secured by combinations of four coils of resistance $4\Omega$, $8\Omega$, $12\Omega$, $24\Omega$?
    Show answer

    (a) Highest: $48\Omega$ with all four in series; (b) lowest: $2\Omega$ with all four in parallel.

    (a) The highest total resistance is obtained with all four coils in series: $R=4+8+12+24=48\Omega$. (b) The lowest is obtained with all four in parallel: $\frac{1}{R}=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24}=\frac{12}{24}=\frac{1}{2}$, so $R=2\Omega$.

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17. Q37 NCERT · Intext Questions (Page 190)
    Why ​does the cord of an electric heater ​not glow while ​the ​heating ​element ​does?
    Show answer

    Both carry the same current, but the element's far higher resistance makes it produce much more heat, so only it glows.

    The cord and the heating element are joined in series, so exactly the same current $I$ flows through both for the same time $t$. By Joule's law the heat produced is $H=I^{2}Rt$, so with $I$ and $t$ common to both, the heat produced in each part is proportional to its resistance. The element is a long, thin coil of nichrome, an alloy of very high resistivity, so its resistance is large (a few tens of ohms). The cord is a short, thick copper wire of very low resistivity, so its resistance is only a small fraction of an ohm. Almost all the heat is therefore produced in the element; its temperature rises until it is red hot and it glows, and nichrome withstands this because it has a high melting point and does not oxidise readily. The heat produced in the cord is negligible, its temperature hardly rises, and so it does not glow.

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18. Q38 NCERT · Intext Questions (Page 190)
    Compute the heat generated while transferring $96000$ coulomb ​of ​charge ​in ​one ​hour ​through a potential difference of $50$ V.
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    The heat generated is $4.8\times 10^6$ J (4.8 MJ), since each coulomb gives up 50 J.

    Each coulomb falls through $50$ V and so gives up $50$ J, and the whole of that energy appears as heat. Therefore $H=VQ=50\text{V}\times 96000\text{C}=4.8\times 10^6$ J $=4.8$ MJ (equal to $1.33$ kW h). The time of one hour is not needed for the heat; it only fixes the current, $I=Q/t=96000/3600=26.67$ A.

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19. Q39 NCERT · Intext Questions (Page 190)
    An ​electric ​iron of resistance $20\Omega$ takes ​a current of $5$ A. Calculate the heat developed ​in $30$ ​s.
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    $H=1.5\times 10^4$ J: the iron develops 15 kJ of heat in 30 s.

    $H=I^{2}Rt=(5\text{A}{)}^2\times 20\Omega \times 30\text{s}=25\times 20\times 30=15000$ J $=1.5\times 10^4$ J. The iron produces heat at the rate $P=I^{2}R=500$ W, so in $30$ s it develops $15$ kJ.

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20. Q40 NCERT · Intext Questions (Page 192)
    What determines the ​rate ​at ​which energy ​is ​delivered by a ​current?
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    Electric power, $P=VI$, set by the potential difference across the device and the current through it.

    The rate at which energy is delivered by a current is the electric power, $P$. It is determined by two things: the potential difference $V$ across the device, which is the energy given up by each coulomb, and the current $I$ through it, which is the number of coulombs passing each second. Hence $P=VI$, measured in watts ($1$ W $=1$ V $\times$ $1$ A $=1$ J/s). Using Ohm's law the same rate can be written $P=I^{2}R=V^2/R$, so for a device of fixed resistance it is settled by the supply voltage, or equally by the current it draws.

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